The emf of a string in a circle - magnetic fields

In summary: In the second question, the electric field on the wire and the resistor creates an emf across the loop.
  • #1
axcelenator
33
0

Homework Statement


there is a cylindrical area in xy plane with a magnetic field (into the plane) that changes with time like that: (dB/dt)=δ
the magnetic field outside the area is 0. the radius of the cylinder is a.
we pick a string in the circle with lentgh L which is smaller than the diameter(2a). let's call the points that connect the string - A and B.
I have to find the emf between those points.

its looks like that:

http://img714.imageshack.us/img714/7092/emfk.jpg

the answer is -(Lδ/2)√[(a^2)-(L/2)^2]
please show me how can I get it. thanks
 
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  • #2
I got it but can't understand something: if i take the inner loop that creates the triangle I'll get the answer. But, if i take the loop that goes from A to B via the arc of the circle the EMF will be other. why?(i understand that the Elec field is nonconservative)
thanks
 
  • #3
You need to show us your calculations.
 
  • #4
Hey Vela, the original question goes like that: there is a conducting rod between A and B and a resistor R connected to it: http://img692.imageshack.us/img692/2079/faraday.jpg

Now. in the first question(when there is no resistor and no conductor) I'm asked to find only the EMF between the points.
I did it like that: i picked the loop that created the triangle and so: EMF=-(d/dt)∫ ∫B*dS=
-δ*(territory of the traingle)=-(Lδ/2)√[(a^2)-(L/2)^2]
(that is the right answer by the teacher)


In the second question the resistor connected to the conductor. What is the current via the resistor?? so I took now the area under the triangle and the new calculation of the flux, EMF and in the end the current gave me:
I=(-δ/R)*{[(a^2)(arcsin(L/2a)]-[(L/2)√[(a^2)-(L/2)^2]]}
(that is the right answer by the teacher)
Now the problem is like that: the teacher asks why can't you say that the current is not:
I=EMF/R=-(Lδ/2R)√[(a^2)-(L/2)^2] ??
Thanks
 
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  • #5
That's a good question. What do you think? You're probably assuming there's no emf induced in the wires connecting the resistor to the rod. Is that true?
 
  • #6
emmmm that's nice.. (-:
I didn't assume like that but I thought:(in the first question) if i took a smaller triange with a smaller area so the flux is smaller too my EMF would be smaller... -->so if i take a loop with an area that limits 0 I would get EMF≈0... LOL (-:
Where is my mistake sir?
 
  • #7
The EMF around the entire loop would be smaller, but in the first part, you're only interested in the EMF along the rod. That wouldn't change because you chose a different loop, but the total EMF around the loop would be affected by the EMF induced on the other parts of the loop, which would change if you modified the loop.
 
  • #8
i can't understand why the current of R is not the EMF/R - (emf of the triangle)
 
  • #9
Alright now I understand it: the emf of the triangle is the emf of the rod only becaude the Electric field on the radius is perpendicular to it and so E*dl = 0.

And in the second question the E*dl on the wires and the resistor is not 0 so I get another EMF across the loop!
 
  • #10
Right!
 

Related to The emf of a string in a circle - magnetic fields

1. What is the emf of a string in a circle?

The emf, or electromotive force, of a string in a circle refers to the induced voltage that is created when a conducting string moves in a circular path within a magnetic field.

2. How is the emf of a string in a circle calculated?

The emf of a string in a circle can be calculated using the equation E = Blv, where B is the strength of the magnetic field, l is the length of the string, and v is the speed at which the string is moving.

3. What factors affect the emf of a string in a circle?

The emf of a string in a circle is affected by the strength of the magnetic field, the length of the string, and the speed at which the string is moving. Additionally, the angle at which the string is moving in relation to the magnetic field also affects the emf.

4. How does the emf of a string in a circle relate to Faraday's law?

The emf of a string in a circle is a direct result of Faraday's law of electromagnetic induction, which states that a changing magnetic field will induce an electric field. In this case, the movement of the string within the magnetic field creates a changing magnetic flux, which induces the emf in the string.

5. What are some real-world applications of the emf of a string in a circle?

The emf of a string in a circle has many practical applications, such as in electric generators, where it is used to convert mechanical energy into electrical energy. It is also used in various instruments, such as electric guitars and speedometers, to measure the rate of movement or rotation.

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